Given three points (-2,2), (-1,-1),(2,6) how do you write a quadratic function in standard form with the points?
1 Answer
f(x) =4/3x^2+x-4/3
Explanation:
Here's an alternative method that I think is quite fun.
Suppose the quadratic function we want to find is
We are told:
f(-2) = 2
f(-1) = -1
f(2) = 6
Also let:
f(0) = p
f(1) = q
Then form a sequence with the values of
2, -1, p, q, 6
Since these are values of a quadratic function for an equally spaced sequence of values of
So starting with:
2, -1, p, q, 6
write down the sequence of differences:
-3, p+1, q-p, 6-q
then the sequence of differences of those differences:
p+4, q-2p-1, p-2q+6
then the sequence of differences of those differences:
q-3p-5, 3p-3q+7
Both of these last two terms must be
{ (q-3p-5 = 0), (3p-3q+7 = 0) :}
If we add these two equations together we find:
-2q+2 = 0
Hence:
q=1
Then from the second equation we find:
3p-3+7 = 0
Hence:
p = -4/3
So our original sequence
2, -1, -4/3, 1, 6
Next write out just the sequence
color(blue)(-4/3), 1, 6
then the sequence of differences:
color(blue)(7/3), 5
then the sequence of differences of differences:
color(blue)(8/3)
We can then use the initial term of each of these sequences to write out a direct formula for the function:
f(x) = color(blue)(-4/3)(1/(0!)) + color(blue)(7/3)(x/(1!)) + color(blue)(8/3)((x(x-1))/(2!))
=-4/3+7/3x+4/3x^2-4/3x
=4/3x^2+x-4/3